Chapter 3-thermodynamics final.docx

7/30/2019 Chapter 3-thermodynamics final.docx

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Thermodynamics I [MIME3110] 22

Chapter 3

THE FIRST LAWOF THERMODYNAMICS

Energy interactions between a system and its surroundings across the boundary in the form

of heat and work have been discussed separately in the previous chapter.

So far, no attempt has been made to relate these interactions between themselves and with the

energy content of the system.

First law of thermodynamics, often called as law of conservation of energy, relating work,

heat, and energy content of the system will be discussed in detail in this chapter.

3.1 First Law of Thermodynamics

In its more general form, the first law may be stated as follows

When energy is either transferred or transformed, the final total energy present in all

forms must precisely equal the original total energy.

It is based on the experimental observations and can not be proved mathematically. All the

observations made so far, confirm the correctness of this law.

3.2 First Law of Thermodynamics for a Closed System

Undergoing a Process

First law can be written for a closed system in an equation form as

systemtheofcontent

energytheinChange

systemthe

leftEnergy

systemthentoi

enteredEnergy

For a system of constant mass, energy can enter or leave the system only in two formsnamely work and heat.

Let a closed system of initial energy E1

receives Q units of net heat and gives out W units

of work during a process. If E2

is energy content at the end of the process as given in Figure 3.1,

applying first law we get


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Q W (E2 E

1) ...(3.1)

Where the total energy content

Internal Energy + Kinetic energy + Potential energy

U +

cg

mC2

2

1+ mgz

The term internal energy usually denoted by the letter U is the energy due to such factors aselectron spin and vibrations, molecular motion and chemical bond.

Kinetic energy term is due to the system movement with a velocity C. For stationarysystems this term will be zero. The term g

cis a constant of value 1 in SI unit. It will be dropped

here after since SI unit is followed throughout the book.

Potential energy term is due to the location of the system in the gravitational field. Itremains constant for a stationary system. The unit of energy in SI is kJ.

3.3 The Thermodynamic Property Enthalpy

Consider a stationary system of fixed mass undergoing a quasi-equilibrium constantpressure process

Applying first law

Q12

1W

2 E

2 E

1

where E2 E

1 (U

2 U

1) + m(C2

2 C

12) + mg(Z

2 Z

1)

U2 U

1since it is a stationary system.

also1W

2 p(V

2 V

1)

p2V

2 p

1V

1

Q12 (p

2V

2 p

1V

1) + (U

2 U

1)

Q12

(U2

+ p2V

2) (U

1+ p

1V

1)

The terms within brackets are all properties depending on the end states. This combinationof properties may be regarded as a single property known as enthalpy. It is usually denoted by

the letter H.

ie H U + pV ...(3.3a)

(or) h u + pv ...(3.3b)


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Where h is specific enthalpy in kJ/kg

u is specific internal energy in kJ/kg and

v is specific volume in m3/kg

3.4 Flow Energy

Flow energy is defined as the energy required to move a mass into the a control volume

against a pressure. Consider a mass of volume V entering into a control volume as given in theFigure 3.2 against a pressure p.

The Flow energy Work done in moving the mass

Force distance

pA dx

p (Adx)

pV ...(3.4)

Therefore, Enthalpy Internal energy + Flow energy

3.5 First Law of Thermodynamics for a Control Volume

Mass simultaneously entering and leaving the system is a very common phenomenon in

most of the engineering applications. Control volume concept is applied to these devices by

assuming suitable control surfaces.

To analyze these control volume problems, conservation of mass and energy concepts are

to be simultaneously considered.

Energy may cross the control surface not only in the form of heat and work but also by

total energy associated with the mass crossing the boundaries. Hence apart from kinetic, potential

and internal energies, flow energy should also be taken into account.


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Conservation of mass

volumecontrol

theofcontentmass

theinchangeNet

volumecontrol

theleaving

massTotal

volumecontrol

theentering

massTotal

Conservation of energy

volumecontrol

theofcontent

theenrgyin

changeNet

volumecontrolthe

leavingmass

thewithassociated

energyTotal

volumecontrolthe

enteringmass

thewithassociated

energyTotal

workand

heatofform

theinboundary

thecrossingenergyNet

...(3.5)

Figure 3.3 First Law of Thermodynamics Applied to a control Volume

As a rate equation, it becomes

CVout

out

in

in EZgC

hmZgC

hmWQ

22

22

...(3.6)

3.6 The Steady-state Flow Process

When a flow process is satisfying the following conditions, it is known as a steady flowprocess.

1. The mass and energy content of the control volume remains constant with time.

2. The state and energy of the fluid at inlet, at the exit and at every point within the control

volume are time independent.

3. The rate of energy transfer in the form of work and heat across the control surface is

constant with time.

Control Surface.

Q

.

W

.

inm.

outm

Control

Volume


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Therefore for a steady flow process

...(3.7)

...(3.7)

also...(3.8)

...(3.9)

For problem of single inlet stream and single outlet stream

gZZ

CChhmWQ 12

2

1

2

212

2)( ...(3.10)

This equation is commonly known as steady flow energy equation (SFEE).

3.7 Application of SFEE

SFEE governs the working of a large number of components used in many engineeringpractices. In this section a brief analysis of such components working under steady flowconditions are given and the respective governing equations are obtained.

3.7.1. Turbines

Turbines are devices used in hydraulic, steam and gas turbine power plants. As the fluid

passesthrough the turbine, work is done on the blades of the turbine which are attached to ashaft. Due to the work given to the blades, the turbine shaft rotates producing work.

Figure 3.4 Schematic Representation of a Turbine

outin mm

0 CVE

022

22

Zg

ChmZg

ChmWQ

out

out

in

in

Mass entering

Mass leaving

Shaft work

Control

Surface


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General Assumptions

1. Changes in kinetic energy of the fluid are negligible

2. Changes in potential energy of the fluid are negligible.

)( 12 hhmWQ ...(3.11)

3.7.2 Compressors

Compressors (fans and blowers) are work consuming devices, where a low-pressure fluid iscompressed by utilising mechanical work. Blades attached to the shaft of the turbine imparts

kinetic energy to the fluid which is later converted into pressure energy.

Figure 3.5 Schematic Representation of a Compresso

General Assumptions

1. Changes in the kinetic energy of the fluid are negligible

2. Changes in the potential energy of the fluid are negligible

Governing Equation

Applying the above equations SFEE becomes

)( 12 hhmWQ ...(3.12)

Mass entering

Mass leaving

Shaft work

Control

Surface


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3.7.3 Pumps

Similar to compressors pumps are also work consuming devices. But pumps handleincompressible fluids, whereas compressors deal with compressible fluids.

General Assumptions

1. No heat energy is gained or lost by the fluids;

2. Changes in kinetic energy of the fluid are negligible.

Governing Equation

gZZhhmW 1212 )( ...(3.13)

As the fluid passes through a pump, enthalpy of the fluid increases, (internal energy of thefluid remains constant) due to the increase in pv (flow energy). Increase in potential energy of

fluid is the most important change found in almost all pump applications.

3.7.4 Nozzles

Nozzles are devices which increase the velocity of a fluid at the expense of pressure. A

typical nozzle used for fluid flow at subsonic* speeds is shown in Figure 3.7.

General Assumptions

1. In nozzles fluids flow at a speed which is high enough to neglect heat lost or gainedas it crosses the entire length of the nozzle. Therefore, flow through nozzles can be

regarded as adiabatic. That is 0.

2. There is no shaft or any other form of work transfer to the fluid or from the fluid;

that is 0.

3. Changes in the potential energy of the fluid are negligible.


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Governing Equation

)(2

02

)(

21

2

1

2

2

2

1

2

212

hhCC

CChh

3.7.5 Diffusers

Diffusers are (reverse of nozzles) devices which increase the pressure of a fluid stream byreducing its kinetic energy.

General Assumptions

Similar to nozzles, the following assumptions hold good for diffusers.

1. Heat lost or gained as it crosses the entire length of the nozzle. Therefore, flow through

nozzles can be regarded as adiabatic. That is 0Q

2. There is no shaft or any other form of work transfer to the fluid or from the fluid;that is 0.

3. Changes in the potential energy of the fluid are negligible

In Out

Control Surface


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Hot fluid out

Cold fluid in

coldfluid out

Hot fluid in

Governing Equation

...(3.14)

3.7.6 Heat Exchangers

Devices in which heat is transferred from a hot fluid stream to a cold fluid stream areknown as heat exchangers.

General Assumptions

1. Heat lost by the hot fluid is equal to the heat gained by the cold fluid.

2. No work transfer across the control volume.

3. Changes in kinetic and potential energies of both the streams are negligible.

Governing Equation

For both hot and cold streams

As per the assumption,

coldhot QQ

The negative sign in the LHS is to represent that heat is going out of the system.

)()( 1221 hhmhhm ch ...(3.15)

2)(

02

)(

22

21

12

2

1

2

212

CChh

CChh

)( 12 hhmQ


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3.7.7 Throttling

A throttling process occurs when a fluid flowing in a line suddenly encounters a restriction in theflow passage. It may be

a plate with a small hole as shown in Figure 3.10 (a)

a valve partially closed as shown in Figure 3.10 (b)

a capillary tube which is normally found in a refrigerator as shown in Figure 3.10 (c)

a porous plug as shown in Figure 3.10 (d)


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2

Path A

Path B

p

V

General assumptions

1. No heat energy is gained or lost by the fluid; ie., 0

2. There is typically some increase in velocity in a throttle, but both inlet and exitkinetic energies are usually small enough to be neglected.

3. There is no means for doing work; ie., 0.

4. Changes in potential energy of the fluid is negligible.

Governing Equation

h2 h

1...(3.16)

Therefore, throttling is an isenthalpic process.

3.8 First Law for a Cyclic Process

In a cyclic process the system is taken through a series of processes and finally returned toits original state. The end state of a cyclic process is identical with the state of the system at the

beginning of the cycle. This is possible if the energy level at the beginning and end of the cyclic

process are also the same. In other words, the net energy change in a cyclic process is zero.

Figure 3.11 First Law for a Cyclic Process

Consider a system undergoing a cycle consisting of two processes A & B as shown in

Figure 3.11 Net energy change

EA

+EB 0 ..(3.17)

(QA W

A) + (Q

B W

B) 0 ...(3.18)


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1

2

Path A

Path B

Path C

p

V

ie QA Q

B W

A W

B...(3.19)

(or) dWdQ ...(3.20)Hence for a cyclic process algebraic sum of heat tranfers is equal to the algebraic sum of

work transfer.

This was first proved by Joule, based on the experiments he conducted between 1843 and1858, that were the first quantitative analysis of thermodynamic systems.

3.9 Energy is a property of a system

Consider a system undergoing a process from state1 to state2 along path A as shown inFigure 3.12. Let the system be taken back to the initial state 1 along two possible paths B and C.Process A, combined separately with process B and C forms two possible cycles.

Figure 3.12 Illustration to show that energy is property

Cycle 1A2B1

QA QB [WA WB]

QA W

A [Q

B W

B]

EAE

B...(3.21)

Cycle 1A2C1

QA Q

C [W

A W

C]


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QA W

A [Q

C W

C]

EA E

C...(3.22)

From Equation (3.21) and (3.22) it can be concluded that energy change in path B andpath C are equal and hence energy is a point function depending only on the end states.

It has been already shown that all the properties are point functions and hence energy isalso a property of the system.

3.10 Specific Heat at Constant Volume and at Constant Pressure

Specific heat at constant volume of a substance is the amount of heat added to rise thetemperature of unit mass of the given substance by 1 degree at constant volume

From first law for a stationary closed system undergoing a process

dQ pdV + dU or dq pdv + du

For a constant volume process

dQ dU or dq du

or du CvdT ...(3.23)

Similarly specific heat at constant pressure is the quantity of heat added to rise the temperature ofunit mass of the given substance by 1 degree at constant pressure

where dQ pdV + dU

dQ pdV + d(H PV)

dQ pdV + dH Vdp pdV

dQ dH Vdp

For a constant pressure process dp 0

Hence dQ dH or dq dh

or dh CpdT ....(3.24)

Note

For solids and liquids, constant volume and constant pressure processes areidentical and hence, there will be only one specific heat.

The difference in specific heats Cp C

v R

The ratio of sp. heat Cp/CvSince h and u are properties of a system, dh C

pdT and duC

vdT, for all

processes.


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3.11 Work Interaction in a Reversible Steady Flow Process

In a steady flow process the work interaction per unit mass between an open system andthe surroundings can be expressed in differential form as

dq dw dh + CdC + gdz

dw dq (dh + CdC +gdz)

Also, dq du + pdv (or) dh vdp

Therefore, dw dh vdp (dh + CdC + gdz)

vdp (CdC + gdz)

For a stationary system

...(3.26)

3.12 First law for an open system under unsteady flow conditions

Many processes of engineering interest involve unsteady flow, where energy andmass content of the control volume increase or decrease.

Example for such conditions are:

1) Filling closed tanks with a gas or liquid.

2) Discharge from closed vessels.

3) Fluid flow in reciprocating equipments during an individual cycle.

To develop a mathematical model for the analysis of such systems the following

assumptions are made.

1) The control volume remains constant relative to the coordinate frame.

2) The state of the mass within the control volume may change with time, but at

any instant of time the state is uniform throughout the entire control volume.

3) The state of the mass crossing each of the areas of flow on the control surfaceis constant with time although the mass flow rates may be time varying.

)(2 12

2

1

2

2

2

1

zzgCC

vdpW

2

1

vdpW


15/20


Unlike in steady flow system, duration of observation t plays an important role intransient analysis. Let mass of the working fluid within the control volume before and after the

observation be m1

and m2

respectively. Applying mass balance we get,

(m2 m

1)

CVm

im

0...(3.27)

Where mi is the mass entered the control volume during the interval t seconds.

m0

is the mass left the control volume during the interval t seconds.

By applying energy balance we get,

...(3.28)

Where ECV

is the change in energy content of the control volume in t seconds.

QCV

is the heat energy entered into the control volume in t seconds.

WCV

is the work energy left the control volume in t seconds.

hi& h

0are specific enthalpy of the inlet and outlet streams respectively.

are the kinetic energy of the inlet and outlet streamsrespectively.

Zig & Z

0g are the potential energy of inlet and outlet streams

respectively.

3.13 Perpetual Motion Machine - IAn engine which could provide work transfer continuously without heat transfer is known as

perpetual motion machine of first kind. It is impossible to have such an engine as it violates firstlaw of thermodynamics.

cvout

out

in

incvcv EZgC

hmZgC

hmWQ

22

22


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Exercises

1. Define internal energy.

2. Express mathematically first law of thermodynamic for the following.

a. a closed system undergoing a process

b. a stationary system of fixed mass undergoing a change of state

c. a closed system undergoing a cycle.

d. an open system.

e. an open system with steady-state flow conditions.

3. Define flow energy and enthalpy.

4. For a stationary system of fixed mass undergoing a process such that its volume remains

constant,

Q12U(T/F)

5. dQ dh vdp for closed system undergoing a process (T/F).

6. Define specific heat at (a) constant pressure (b) constant volume

7. Determine the power of the cycle comprising four processes in which the heat transfers are

: 50 kJ/kg, 20 kJ/kg, 7l J/kg and 12 kJ/kg having 100 cycles per minute.

[48.3 kW]

8. Write the steady flow energy equation and explain the terms involved in it.

9. Show that energy is a property of the system.

10. What are conditions for steady flow process?

11. A piston-cylinder assembly contains 1kg or nitrogen at 100 kPa. The initial volume is 0.5

m3. Heat is transferred to the substance in an amount necessary to cause a slow expansion

at constant temperature. This process is terminated when the final volume is twice the

initial volume.

[34.7 kJ]

12. 2 kg of air enclosed in a rigid container receives 0.2 kJ of paddle wheel work and 0.5 kJ of

electrical energy per second. Heat loss from the system is 0.6 kJ/s. If the initial

temperature is 25oC what will be the temperature after 5 minutes?

[45.9o

C]

13. A well insulated, frictionless piston-cylinder assembly contains 0.5 kg of air initially at

75oC and 300 kPa. An electric - resistance heating element inside the cylinder is energized

and causes the air temperature to reach 150oC. The pressure of the air is maintained

constant throughout the process. Determine the work for the process and the amount of

electrical work.

{Hint Qnet Wnet =U; Wnet=+Welectric}


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[26.9 kJ ; 37.7]

14. A cylinder contains 168 litres of a gas at a pressure of 1 bar and temperature of 47oC. If

this gas is compressed to one-twelfth of its volume, pressure is then 21 bar. Find

a. index of compressionb. change in internal energy

c. heat rejected during compression

Take Cp 1.089 and Cv 0.837 both in kJ/kg

[1.225 ; 41.81 kJ ; 14.05 kJ]15. a. A mass of 10 kg is falling from a height of 100 m from the datum. What will be the

velocity when it reaches a height of 20 m from the datum? Take the total heat loss from

the mass when it falls from 100 m height to 20 m height is 5 kJ.

[8.68 m/s]b. An insulated box containing carbon dioxide gas falls from a balloon 3.5 km above the

earths surface. Determine the temperature rise of the carbon dioxide when box hits the

ground.

Take Cv 0.6556 kJ/kg

[52.37oC]

16. A working substance flows at a rate of 5 kg/s into a steady flow system at 6 bar, 2000 kJ/kg

of internal energy and 0.4 m3/kg specific volume with a velocity of 300 m/s. It leaves at 10

bar, 1600 kJ/kg internal energy, 1.2 m3/kg specific volume with a velocity of 150 m/s. The

inlet is 10m above the outlet. The work transfer to the surroundings in 3 MW. Estimate the

heat transfer and indicate the direction.

[5630 kJ/s]

17. An air compressor takes in air at 100 kPa, 40oC and discharges it at 690 kPa, 208

oC. The

initial and final internal energy values for the air are 224 and 346 kJ/kg respectively. The

cooling water around the cylinders removes 70 kJ/kg from the air. Neglecting changes in

kinetic and potential energy, calculate the work.

[100.216 kJ/kg]

18. A perfect gas of cp 1.1 kJ/kg flows through a turbine at a rate of 3 kg/s. The inlet and exit

velocity are 30 and 130 m/s respectively. The initial and final temperature are 650

o

C and250oC respectively. Heat loss is 45 kJ/s. Find the power developed.

[1251 kW]

19. In a turbine 4500 kg/min of air expands polytropically from 425 kPa and 1360 K to 101

kPa. The exponent n it equal to 1.45 for the process. Find the work and heat.

[33939 kW ; 2927 kJ/s]


18/20


20. Air expands through a nozzle from a pressure of 500 kPa to a final pressure of 100 kPa.

The enthalpy decrease by 100 kJ/kg. The flow is adiabatic and the inlet velocity is very

low. Calculate the exit velocity.

[447.2 m/s]

21. A closed system undergoes a cycle consisting of three process 1-2, 2-3 and 3-1. Given thatQ12 30 kJ, Q23 10 kJ, 1w2 5 kJ, 3w2 5 kJ and E31 15 kJ, determine Q31, w23,E12

and E23.

[20 kJ ; 50 kJ ; 25 kJ ; 40 kJ ]22. The following cycle involves 3 kg of air : Polytropic compression from 1 to 2 where P1

150 kPa, T1 360 K, P2 750 kPa and n 1.1 ; constant-pressure cooling from 2 to 3; and

constant - temperature heating from 3 to 1. Draw the pV diagram and find temperatures,

pressures and volumes at each state and determine the net work and heat.

[150 kPa ; 2.066 m3; 360 K ; 750 kPa ; 0.478 m

3;

416.72 K ; 750 kPa ; 0.414 m3

; 360 K ; 35 kJ]23. A cycle, composed of three processes, is :

Polytropic compression (n 1.5) from 137 kPa and 38oC to state 2 ; constant pressure

process from state 2 to state 3 ; constant volume process form state 3 and to state 1. The

heat rejected in process 3-1 is 1560 kJ/kg and the substance is air. Determine

(a) the pressures, temperatures and specific volumes around the cycle

(b) the heat transfer in process 1-2

(c) the heat transfer in process 2-3

(d) work done in each process and(e) net work done in the cycle

[137 kPa ; 0.6515 m3/kg ; 311.0 K ; 1095 kPa ; 0.1630 m

3/kg ;

621.8 K ; 1095 kPa ; 0.6515 m3/kg ; 2487.0 K ; 44.44 kJ ;

1872.25 kJ ; 178 kJ ; 534.9 kJ ; 0 ; 356.9 kJ]24. 0.15 m

3of air at a pressure of 900 kPa and 300

oC is expanded at constant pressure to 3

time its initial volume. It is then expanded polytropically following the lawPV1.5

C and

finally compressed back to initial state isothermally. Calculate

(a) heat received

(b) heat rejected(c) efficiency of the cycle

[944.5kJ ; 224.906 kJ ; 0.291]25. A piston and cylinder device contains 1 kg of air, Initially, v 0.8 m

3/kg and

T 298 K. The air is compressed in a slow frictionless process to a specific volume of 0.2

m3/kg and a temperature of 580 K according to the equation pV

1.3 0.75 ( p in bar, v in

m3/kg). If Cv of air is 0.78 kJ/kg determine :


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(a) work and

(b) heat transfer (both in kJ)

[ 137.85 kJ ; 82.11 kJ]26. The internal energy of a closed system is given by U 100 + 50 T + 0.04 T

2in Joules, and

the heat absorbed by Q 4000 + 16 T in Joules, where T is in Kelvin. If the systemchanges from 500 K to 1000 K, what is the work done ?

[47 kJ]

27. One kg of air, volume 0.05 m3, pressure 20 bar expands reversibly according to the law

pv1.3

C until the volume is doubled. It is then cooled at constant pressure to initial volumeand further heat at constant volume so that it returns back to initial process. Calculate the

network done by air.

[21.98 kJ]

28. Air at the rate of 14 kg/s expands from 3 bar, 150C to 1bar reversibly andadiabatically. Find the exit temperature and power developed. Neglect thechanges in kinetic and potential energy. [ 309 k ; 1.603 kW]

29. Specific internal energy of a certain substance can be expressed as follows:

u 831.0 + 0.617 pv

Where u is the specific internal energy in kJ/kg

p is the pressure in k Pa

v is the specific volume in m3/kg

One kg of such substance expands from 850 kPa, 0.25 m3/kg to 600 kPa,

0.5 m3 /kg. Find the work done and heat transferred. [ 176.06 kJ ; 230 kJ]

30. A cylinder of 8 cm internal diameter is fitted with a piston loaded by a coil springof stiffness 140 N/cm of compression. The cylinder contains 0.0005 m3 of air at

15C and 3 bar. Find the amount of heat which must be supplied for the piston toa distance of 4 cm. Sketch the process on a p-V diagram.

[ 0.417 kJ]

31. Prove that

Q mCv

for a polytropic process of index n.

32. An air conditioning system for a computer room in a tower block draws in air on the roof at

a height of 100 m with a velocity of 25 m/s. The air is at 28oC. The air is discharged at a

height of 10 m with a velocity of 2 m/s at 14oC. The mass flow rate is 2 kg/s, and a heat

transfer of 40.73 kW cools the air before it is discharged. Calculate the rate of work for

the air passing through the system. Take Cp for air as 1005 J/kgK.

[ 10.23 kW]33. A diffuser reduces the velocity of an air stream from 300 m/s to 30 m/s. If the inlet pressure

and temperature are 1.01 bar and 315oC, determine the outlet pressure. Find also the area

required for the diffuser to pass a mass flow of 9 kg/s.


20/20

Th d i I [MIME3110] 41

[4.586 bar, 0.17 m2]

34. A centrifugal air compressor operating at steady state has an air intake of 1.2 kg/min. Inlet

and exit conditions are as follows:

Properties p (kPa) ToC u kJ/kg v m

3/kg

Inlet 100 0 195.14 0.784Exit 200 50 230.99 0.464

If the heat loss is negligible, find the power input. [ 1.005 kW]

35. A household gas cylinder initially evacuated is filled by 15 kg gas supply of enthalpy 625

kJ/kg. After filling, the gas in the cylinder has the following parameters :

pressure 10 bar ;

enthalpy 750 kJ/kg and

specific volume 0.0487 m3/kg.

Evaluate the heat received by the cylinder from the surroundings.

[1144.5 kJ]

36. 0.56 m3

of air at 0.2 MPa is contained in a fully insulated rigid vessel. The vessel

communicates through a valve with a pipe line carrying high pressure air at 300 K

temperature. The valve is opened and the air is allowed to flow into the tank until the

pressure of air in the tank is raised to 1MPa. Determine the mass of air that enters the tank.

Neglect kinetic energy of the incoming air.

[3.72 kg]

37. An insulated rigid tank contains 8 kg of air at 1.5 bar pressure and 310 K temperature. It is

filled with air from a large reservoir at 15 bar and 335 K. If the air behaves as a perfect gas,

make calculations for the amount of air added and its temperature.

[47.6 kg ; 446.04K]

38. A pressure vessel contains a gas at an initial pressure of 3.5 MN/m2

and at a temperature of

60oC. It is connected through a valve to a vertical cylinder in which there is a piston. The

valve is opened, gas enters the vertical cylinder, and work is done in lifting the piston. The

valve is closed and the pressure and the temperature of the remaining gas in the cylinder

are 1.7 MN/m2

and 25oC, respectively. Determine the temperature of the gas in the vertical

cylinder if the process is assumed to be adiabatic. Take 1.4.

[267.6 K]

39. A pressure vessel is connected, via a valve, to a gas main in which a gas is maintained at a

constant pressure and temperature of 1.4 MN/m2

and 85oC, respectively. The pressure

vessel is initially evacuated. The valve is opened and a mass of 2.7 kg of gas passes intothe pressure vessel. The valve is closed and the pressure and temperature of the gas in the

pressure vessel are then 700 KN/m2

and 60oC, respectively. Determine the heat transfer to

or from the gas in the vessel. Determine the volume of the vessel and the volume of the gas

before transfer.

For the gas, take Cp 0.88 kJ/kgK, Cv 0.67 . Neglect velocity of the gas in the main

[248.2 kJ ; 0.27 m3 ; 0.145 m3 ]

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Chapter 3-thermodynamics final.docx